<?xml version="1.0" encoding="UTF-8"?>

<record version="2" id="643">
 <title>inflexion point</title>
 <name>InflexionPoint</name>
 <created>2009-04-17 19:11:18</created>
 <modified>2026-09-09 20:40:46</modified>
 <type>Definition</type>
 <creator id="21" name="pahio"/>
 <modifier id="1" name="bloftin"/>
 <comment>commenting out picture for now until correction</comment>
 <author id="21" name="pahio"/>
 <classification>
	<category scheme="pacs" code="02.30.-f"/>
 </classification>
 <preamble>% Object 643: Inflexion Point
% PSTricks/pst-plot removed; use a static PNG for the live article.
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
</preamble>
 <content>In examining the graphs of differentiable real functions, it may be useful to
\PMlinkescapetext{state} the intervals where the function is
\PMlinkescapetext{convex} and those where it is
\PMlinkescapetext{concave}.

\begin{itemize}

\item A function $f$ is said to be
\PMlinkescapetext{\emph{convex} on an interval} if the
\PMlinkname{restriction}{RestrictionOfAFunction} of $f$ to that interval is
convex. Geometrically, the graph is often described as
\PMlinkescapetext{\emph{concave upward}}. If $f$ is twice differentiable,
$f''(x)\geq 0$ throughout the interval is sufficient for convexity. For the
sine curve, $f''(x)=-\sin x&gt;0$ on $(-\pi,0)$.

\item Correspondingly, $f$ is concave downward on an interval when its graph
bends downward. For a twice differentiable function, $f''(x)\leq 0$ throughout
the interval is sufficient for concavity. For the sine curve,
$f''(x)=-\sin x&lt;0$ on $(0,\pi)$.

\item A point at which the graph changes from concave upward to concave
downward, or vice versa, is called an \emph{inflection point}
(or \emph{inflexion point}). For a twice differentiable function, an
inflection point often satisfies $f''(x)=0$, but the essential condition is
that the concavity changes sign across the point.

\end{itemize}

\begin{center}
\includegraphics{Inflexion_Point_Sine.png}

\emph{The origin is an inflection point of the sinusoid $y=\sin x$.}
\end{center}

Since the sine function is $2\pi$-periodic, the sinusoid possesses infinitely
many inflection points. Indeed,
\[
f(x)=\sin x,
\qquad
f''(x)=-\sin x.
\]
Thus
\[
f''(x)=0
\]
for
\[
x=n\pi,
\qquad
n\in\mathbb{Z}.
\]
Moreover,
\[
f'''(x)=-\cos x,
\]
so
\[
f'''(n\pi)
=
-\cos(n\pi)
=
(-1)^{n+1}
\neq 0.
\]
Hence $f''$ changes sign at every $x=n\pi$, confirming that these points are
inflection points.

\textbf{Remarks}

1. For finding the inflection points of the graph of $f$, it does not suffice
merely to find the \PMlinkname{roots}{Equation} of
\[
f''(x)=0,
\]
because the sign of $f''$ need not change as such a root is crossed. For
example, $x=0$ is not an inflection point of $f(x)=x^4$, since
\[
f''(x)=12x^2
\]
does not change sign at the origin.

2. Recalling that the signed \PMlinkname{curvature}{CurvaturePlaneCurve} of a
plane curve $y=f(x)$ may be written
\[
\kappa(x)
=
\frac{f''(x)}
{\left[1+f'(x)^2\right]^{3/2}},
\]
an inflection point is associated with a change in the sign of the signed
curvature.

3. If an inflection point $x=\xi$ also satisfies
\[
f'(\xi)=0,
\]
it is called a \PMlinkescapetext{\emph{stationary inflection point}}. If
\[
f'(\xi)\neq 0,
\]
it is a non-stationary inflection point.
</content>
</record>
